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A stable linear noise covariance balance has a positive solution

Example

For $A = diag (- 1, - 2)$ and noise-factor column $B = (1, 1)^{T}$ , the stable covariance-balance equation is $A X + X A^{T} + B B^{T} = 0$ . Its unique positive solution is

X = (\begin{array}{cc} 1 / 2 & 1 / 3 \\ 1 / 3 & 1 / 4 \end{array}) .

This verifies the algebraic covariance balance and its positive solution; it does not assert a separate stochastic realization or stationary-law theorem.

Facts & Assumptions

Given: The matrices in the Example.

[F1]

Hurwitz matrices and the continuous Lyapunov equation: $A$ is Hurwitz when all its eigenvalues have negative real part, and the balance is $A X + X A^{T} + Q = 0$ .

[F2]

The Hurwitz Lyapunov integral is the unique solution and has the stated positivity: For Hurwitz $A$ , the Lyapunov equation has a unique solution, and the integral solution is positive definite under the stated conditions.

Verification

1.1givenF1algebra

The eigenvalues of $A$ are $- 1$ and $- 2$ , so $A$ is Hurwitz by [F1]. Direct multiplication gives $B B^{T} = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ and $A X + X A^{T} = (\begin{array}{cc} - 1 & - 1 \\ - 1 & - 1 \end{array}) = - B B^{T}$ . Thus the displayed $X$ solves the balance.

2.1step 1.1algebra

For real $(u, v)$ , $(\begin{array}{cc} u & v \end{array}) X (\begin{array}{c} u \\ v \end{array}) = \frac{1}{2} (u + \frac{2}{3} v)^{2} + \frac{1}{36} v^{2}$ , which is positive unless $u = v = 0$ . Hence $X ≻ 0$ .

3.1F1F2step 1.1step 2.1∎

Since $A$ is Hurwitz, [F2] gives uniqueness of the solution, so this positive matrix is the unique stable Lyapunov covariance-balance solution.

Depends on

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